A series LCR circuit resonates at f₀ = 1000 Hz and its half-power frequencies are 990 Hz and 1010 Hz. Find (a) the bandwidth, (b) the Q factor, (c) the voltage across the inductor at resonance if the source is 2 V rms. Step 1: Bandwidth Δf = f₂ − f₁ = 1010 − 990 = 20 Hz. (In angular terms Δω = 2π × 20 rad/s, but the ratio Q = f₀/Δf works the same.) Step 2: Q = f₀/Δf = 1000/20 = 50. This is a fairly sharp resonance — the circuit responds strongly only within ±1% of 1000 Hz. Step 3: At resonance, V_L = Q × V_source = 50 × 2 = 100 V rms across the inductor (and 100 V across the capacitor, opposite in phase so they cancel). The source supplies only 2 V but the coil sees 100 V — this voltage magnification is exactly what Q measures, and why high-Q power circuits need components rated for large voltages.